mathop {lim }limits_{y to 0} frac{(x + y)sec | vedclass

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Question

(A) Let $f(y) = (x + y)\sec(x + y)$. We need to find $\mathop {\lim }\limits_{y 	o 0} \frac{f(y) - f(0)}{y}$,which is the definition of the derivativ

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$\sec x(x	an x + 1)$

Vedclass · Answer

(A) Let $f(y) = (x + y)\sec(x + y)$. We need to find $\mathop {\lim }\limits_{y 	o 0} \frac{f(y) - f(0)}{y}$,which is the definition of the derivative $f'(0)$ with respect to $y$.$f'(y) = \frac{d}{dy} [(x + y)\sec(x + y)]$Using the product rule: $f'(y) = 1 \cdot \sec(x + y) + (x + y) \cdot \sec(x + y)	an(x + y)$.Evaluating at $y = 0$:$f'(0) = \sec(x) + x\sec(x)	an(x)$$f'(0) = \sec x(1 + x	an x)$.

$\mathop {\lim }\limits_{y \to 0} \frac{(x + y)\sec (x + y) - x\sec x}{y} = $

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